We can solve the problem by using the first law of thermodynamics:
[tex]\Delta U = Q-W[/tex]
where
[tex]\Delta U[/tex] is the variation of internal energy of the system
Q is the heat absorbed by the system
W is the work done by the system
In this problem, the heat absorbed by the system is Q=+44 J, and the increase in internal energy is [tex]\Delta U = +32 J[/tex]. So we can rearrange the equation to calculate the work done:
[tex]W=Q-\Delta U=44 J-32 J=+12 J[/tex]
and the work is positive, which means that it is done by the system on the surrounding.
